Given a sorted array arr[] and a value X, find the k closest elements to X in arr[].
Examples:
Input: K = 4, X = 35
arr[] = {12, 16, 22, 30, 35, 39, 42,
45, 48, 50, 53, 55, 56}
Output: 30 39 42 45
Note that if the element is present in array, then it should not be in output, only the other closest elements are required.
In the following solutions, it is assumed that all elements of array are distinct.
A simple solution is to do linear search for k closest elements.
- Start from the first element and search for the crossover point (The point before which elements are smaller than or equal to X and after which elements are greater). This step takes O(n) time.
- Once we find the crossover point, we can compare elements on both sides of crossover point to print k closest elements. This step takes O(k) time.
The time complexity of the above solution is O(n).
An Optimized Solution is to find k elements in O(Logn + k) time. The idea is to use Binary Search to find the crossover point. Once we find index of crossover point, we can print k closest elements in O(k) time.
#include<stdio.h> /* Function to find the cross over point (the point before which elements are smaller than or equal to x and after which greater than x)*/ int findCrossOver( int arr[], int low, int high, int x)
{ // Base cases if (arr[high] <= x) // x is greater than all
return high;
if (arr[low] > x) // x is smaller than all
return low;
// Find the middle point int mid = (low + high)/2; /* low + (high - low)/2 */
/* If x is same as middle element, then return mid */ if (arr[mid] <= x && arr[mid+1] > x)
return mid;
/* If x is greater than arr[mid], then either arr[mid + 1] is ceiling of x or ceiling lies in arr[mid+1...high] */
if (arr[mid] < x)
return findCrossOver(arr, mid+1, high, x);
return findCrossOver(arr, low, mid - 1, x);
} // This function prints k closest elements to x in arr[]. // n is the number of elements in arr[] void printKclosest( int arr[], int x, int k, int n)
{ // Find the crossover point
int l = findCrossOver(arr, 0, n-1, x);
int r = l+1; // Right index to search
int count = 0; // To keep track of count of elements already printed
// If x is present in arr[], then reduce left index
// Assumption: all elements in arr[] are distinct
if (arr[l] == x) l--;
// Compare elements on left and right of crossover
// point to find the k closest elements
while (l >= 0 && r < n && count < k)
{
if (x - arr[l] < arr[r] - x)
printf ( "%d " , arr[l--]);
else
printf ( "%d " , arr[r++]);
count++;
}
// If there are no more elements on right side, then
// print left elements
while (count < k && l >= 0)
printf ( "%d " , arr[l--]), count++;
// If there are no more elements on left side, then
// print right elements
while (count < k && r < n)
printf ( "%d " , arr[r++]), count++;
} /* Driver program to check above functions */ int main()
{ int arr[] ={12, 16, 22, 30, 35, 39, 42,
45, 48, 50, 53, 55, 56};
int n = sizeof (arr)/ sizeof (arr[0]);
int x = 35, k = 4;
printKclosest(arr, x, 4, n); return 0;
} |
// Java program to find k closest elements to a given value class KClosest
{ /* Function to find the cross over point (the point before
which elements are smaller than or equal to x and after
which greater than x)*/
int findCrossOver( int arr[], int low, int high, int x)
{
// Base cases
if (arr[high] <= x) // x is greater than all
return high;
if (arr[low] > x) // x is smaller than all
return low;
// Find the middle point
int mid = (low + high)/ 2 ; /* low + (high - low)/2 */
/* If x is same as middle element, then return mid */
if (arr[mid] <= x && arr[mid+ 1 ] > x)
return mid;
/* If x is greater than arr[mid], then either arr[mid + 1]
is ceiling of x or ceiling lies in arr[mid+1...high] */
if (arr[mid] < x)
return findCrossOver(arr, mid+ 1 , high, x);
return findCrossOver(arr, low, mid - 1 , x);
}
// This function prints k closest elements to x in arr[].
// n is the number of elements in arr[]
void printKclosest( int arr[], int x, int k, int n)
{
// Find the crossover point
int l = findCrossOver(arr, 0 , n- 1 , x);
int r = l+ 1 ; // Right index to search
int count = 0 ; // To keep track of count of elements
// already printed
// If x is present in arr[], then reduce left index
// Assumption: all elements in arr[] are distinct
if (arr[l] == x) l--;
// Compare elements on left and right of crossover
// point to find the k closest elements
while (l >= 0 && r < n && count < k)
{
if (x - arr[l] < arr[r] - x)
System.out.print(arr[l--]+ " " );
else
System.out.print(arr[r++]+ " " );
count++;
}
// If there are no more elements on right side, then
// print left elements
while (count < k && l >= 0 )
{
System.out.print(arr[l--]+ " " );
count++;
}
// If there are no more elements on left side, then
// print right elements
while (count < k && r < n)
{
System.out.print(arr[r++]+ " " );
count++;
}
}
/* Driver program to check above functions */
public static void main(String args[])
{
KClosest ob = new KClosest();
int arr[] = { 12 , 16 , 22 , 30 , 35 , 39 , 42 ,
45 , 48 , 50 , 53 , 55 , 56
};
int n = arr.length;
int x = 35 , k = 4 ;
ob.printKclosest(arr, x, 4 , n);
}
} /* This code is contributed by Rajat Mishra */ |
# Function to find the cross over point # (the point before which elements are # smaller than or equal to x and after # which greater than x) def findCrossOver(arr, low, high, x) :
# Base cases
if (arr[high] < = x) : # x is greater than all
return high
if (arr[low] > x) : # x is smaller than all
return low
# Find the middle point
mid = (low + high) / / 2 # low + (high - low)// 2
# If x is same as middle element,
# then return mid
if (arr[mid] < = x and arr[mid + 1 ] > x) :
return mid
# If x is greater than arr[mid], then
# either arr[mid + 1] is ceiling of x
# or ceiling lies in arr[mid+1...high]
if (arr[mid] < x) :
return findCrossOver(arr, mid + 1 , high, x)
return findCrossOver(arr, low, mid - 1 , x)
# This function prints k closest elements to x # in arr[]. n is the number of elements in arr[] def printKclosest(arr, x, k, n) :
# Find the crossover point
l = findCrossOver(arr, 0 , n - 1 , x)
r = l + 1 # Right index to search
count = 0 # To keep track of count of
# elements already printed
# If x is present in arr[], then reduce
# left index. Assumption: all elements
# in arr[] are distinct
if (arr[l] = = x) :
l - = 1
# Compare elements on left and right of crossover
# point to find the k closest elements
while (l > = 0 and r < n and count < k) :
if (x - arr[l] < arr[r] - x) :
print (arr[l], end = " " )
l - = 1
else :
print (arr[r], end = " " )
r + = 1
count + = 1
# If there are no more elements on right
# side, then print left elements
while (count < k and l > = 0 ) :
print (arr[l], end = " " )
l - = 1
count + = 1
# If there are no more elements on left
# side, then print right elements
while (count < k and r < n) :
print (arr[r], end = " " )
r + = 1
count + = 1
# Driver Code if __name__ = = "__main__" :
arr = [ 12 , 16 , 22 , 30 , 35 , 39 , 42 ,
45 , 48 , 50 , 53 , 55 , 56 ]
n = len (arr)
x = 35
k = 4
printKclosest(arr, x, 4 , n)
# This code is contributed by Ryuga |
// C# program to find k closest elements to // a given value using System;
class GFG {
/* Function to find the cross over point
(the point before which elements are
smaller than or equal to x and after which
greater than x)*/
static int findCrossOver( int []arr, int low,
int high, int x)
{
// Base cases
// x is greater than all
if (arr[high] <= x)
return high;
// x is smaller than all
if (arr[low] > x)
return low;
// Find the middle point
/* low + (high - low)/2 */
int mid = (low + high)/2;
/* If x is same as middle element, then
return mid */
if (arr[mid] <= x && arr[mid+1] > x)
return mid;
/* If x is greater than arr[mid], then
either arr[mid + 1] is ceiling of x or
ceiling lies in arr[mid+1...high] */
if (arr[mid] < x)
return findCrossOver(arr, mid+1,
high, x);
return findCrossOver(arr, low, mid - 1, x);
}
// This function prints k closest elements
// to x in arr[]. n is the number of
// elements in arr[]
static void printKclosest( int []arr, int x,
int k, int n)
{
// Find the crossover point
int l = findCrossOver(arr, 0, n-1, x);
// Right index to search
int r = l + 1;
// To keep track of count of elements
int count = 0;
// If x is present in arr[], then reduce
// left index Assumption: all elements in
// arr[] are distinct
if (arr[l] == x) l--;
// Compare elements on left and right of
// crossover point to find the k closest
// elements
while (l >= 0 && r < n && count < k)
{
if (x - arr[l] < arr[r] - x)
Console.Write(arr[l--]+ " " );
else
Console.Write(arr[r++]+ " " );
count++;
}
// If there are no more elements on right
// side, then print left elements
while (count < k && l >= 0)
{
Console.Write(arr[l--]+ " " );
count++;
}
// If there are no more elements on left
// side, then print right elements
while (count < k && r < n)
{
Console.Write(arr[r++] + " " );
count++;
}
}
/* Driver program to check above functions */
public static void Main()
{
int []arr = {12, 16, 22, 30, 35, 39, 42,
45, 48, 50, 53, 55, 56};
int n = arr.Length;
int x = 35;
printKclosest(arr, x, 4, n);
}
} // This code is contributed by nitin mittal. |
<script> // JavaScript program to find k // closest elements to a given value // Function to find the cross over point // (the point before which elements are // smaller than or equal to x and after // which greater than x) function findCrossOver(arr, low, high, x)
{ // Base cases
if (arr[high] <= x) // x is greater than all
return high
if (arr[low] > x) // x is smaller than all
return low
// Find the middle point
var mid = (low + high) // 2 // low + (high - low)// 2
// If x is same as middle element,
// then return mid
if (arr[mid] <= x && arr[mid + 1] > x)
return mid
// If x is greater than arr[mid], then
// either arr[mid + 1] is ceiling of x
// or ceiling lies in arr[mid+1...high]
if (arr[mid] < x)
return findCrossOver(arr, mid + 1, high, x)
return findCrossOver(arr, low, mid - 1, x)
} // This function prints k closest elements to x // in arr[]. n is the number of elements in arr[] function printKclosest(arr, x, k, n)
{ // Find the crossover point
var l = findCrossOver(arr, 0, n - 1, x)
var r = l + 1 // Right index to search
var count = 0 // To keep track of count of
// elements already printed
// If x is present in arr[], then reduce
// left index. Assumption: all elements
// in arr[] are distinct
if (arr[l] == x)
l -= 1
// Compare elements on left and right of crossover
// point to find the k closest elements
while (l >= 0 && r < n && count < k)
{
if (x - arr[l] < arr[r] - x)
{
document.write(arr[l] + " " )
l -= 1
}
else
{
document.write(arr[r] + " " )
r += 1
}
count += 1
}
// If there are no more elements on right
// side, then print left elements
while (count < k && l >= 0)
{
print(arr[l])
l -= 1
count += 1
}
// If there are no more elements on left
// side, then print right elements
while (count < k && r < n)
{
print(arr[r])
r += 1
count += 1
}
} // Driver Code var arr = [ 12, 16, 22, 30, 35, 39, 42,
45, 48, 50, 53, 55, 56 ]
var n = arr.length
var x = 35
var k = 4
printKclosest(arr, x, 4, n) // This code is contributed by AnkThon </script> |
<?php // PHP Program to Find k closest // elements to a given value /* Function to find the cross over point (the point before
which elements are smaller
than or equal to x and after
which greater than x) */
function findCrossOver( $arr , $low ,
$high , $x )
{ // Base cases
// x is greater than all
if ( $arr [ $high ] <= $x )
return $high ;
// x is smaller than all
if ( $arr [ $low ] > $x )
return $low ;
// Find the middle point
/* low + (high - low)/2 */
$mid = ( $low + $high )/2;
/* If x is same as middle
element, then return mid */
if ( $arr [ $mid ] <= $x and
$arr [ $mid + 1] > $x )
return $mid ;
/* If x is greater than arr[mid],
then either arr[mid + 1] is
ceiling of x or ceiling lies
in arr[mid+1...high] */
if ( $arr [ $mid ] < $x )
return findCrossOver( $arr , $mid + 1,
$high , $x );
return findCrossOver( $arr , $low ,
$mid - 1, $x );
} // This function prints k // closest elements to x in arr[]. // n is the number of elements // in arr[] function printKclosest( $arr , $x , $k , $n )
{ // Find the crossover point
$l = findCrossOver( $arr , 0, $n - 1, $x );
// Right index to search
$r = $l + 1;
// To keep track of count of
// elements already printed
$count = 0;
// If x is present in arr[],
// then reduce left index
// Assumption: all elements
// in arr[] are distinct
if ( $arr [ $l ] == $x ) $l --;
// Compare elements on left
// and right of crossover
// point to find the k
// closest elements
while ( $l >= 0 and $r < $n
and $count < $k )
{
if ( $x - $arr [ $l ] < $arr [ $r ] - $x )
echo $arr [ $l --], " " ;
else
echo $arr [ $r ++], " " ;
$count ++;
}
// If there are no more
// elements on right side,
// then print left elements
while ( $count < $k and $l >= 0)
echo $arr [ $l --], " " ; $count ++;
// If there are no more
// elements on left side,
// then print right elements
while ( $count < $k and $r < $n )
echo $arr [ $r ++]; $count ++;
} // Driver Code $arr = array (12, 16, 22, 30, 35, 39, 42,
45, 48, 50, 53, 55, 56);
$n = count ( $arr );
$x = 35; $k = 4;
printKclosest( $arr , $x , 4, $n );
// This code is contributed by anuj_67. ?> |
39 30 42 45
Time complexity: O(Logn + k).
Auxiliary Space: O(1), since no extra space has been used.
Approach 2: Using Priority Queue
This approach uses a priority queue (max heap) to maintain the k closest numbers to x. It iterates over the elements in the array and calculates their absolute differences from x. The pairs of absolute differences and negative values are pushed into the max heap. If the size of the max heap exceeds k, the element with the maximum absolute difference is removed. Finally, the top k elements from the max heap are extracted and stored in a result vector. The vector is then reversed to obtain the closest numbers in ascending order before being returned as the result.
Below is the implementation:
#include <bits/stdc++.h> using namespace std;
vector< int > findClosestElements(vector< int >& arr, int k,
int x)
{ // Create a max heap to store the pairs of absolute
// differences and negative values
priority_queue<pair< int , int > > maxH;
int n = arr.size();
for ( int i = 0; i < n; i++) {
// Skip if the element is equal to x
if (arr[i] == x)
continue ;
// Calculate the absolute difference and add the
// pair to the max heap
maxH.push({ abs (arr[i] - x), -arr[i] });
// If the size of the max heap exceeds k, remove the
// element with the maximum absolute difference
if (maxH.size() > k)
maxH.pop();
}
// Store the result in a vector
vector< int > result;
// Retrieve the top k elements from the max heap
while (!maxH.empty()) {
// Get the top element from the max heap
auto p = maxH.top();
maxH.pop();
// Add the negative value to the result vector
result.push_back(-p.second);
}
// Reverse the result vector to get the closest numbers
// in ascending order
reverse(result.begin(), result.end());
return result;
} int main()
{ vector< int > arr = { 12, 16, 22, 30, 35, 39, 42,
45, 48, 50, 53, 55, 56 };
int k = 4, x = 35;
vector< int > res = findClosestElements(arr, k, x);
for ( int i = 0; i < res.size(); i++) {
cout << res[i] << " " ;
}
cout << endl;
return 0;
} |
/*package whatever //do not write package name here */ import java.io.*;
import java.util.*;
class GFG {
// Create pair class which implements Comparable
// interface
static class Pair implements Comparable<Pair> {
int absDiff;
int ind;
Pair( int f, int s)
{
absDiff = f;
ind = s;
}
public int compareTo(GFG.Pair o)
{
// If there are two elements with the same
// difference with X, the greater element is
// given priority.
if (absDiff == o.absDiff)
return ind - o.ind;
else
return o.absDiff - absDiff;
}
}
static int [] printKClosest( int [] nums, int n, int k,
int x)
{
PriorityQueue<Pair> maxHeap = new PriorityQueue<>();
for ( int i = 0 ; i < nums.length; i++) {
int diff = Math.abs(nums[i] - x);
//if nums[i] == x then no need to consider that element
if (diff != 0 )
maxHeap.add( new Pair(diff, i));
//if maxheap size exceeds k then remove the element with maximum absolute difference
if (maxHeap.size() > k)
maxHeap.poll();
}
int ans[] = new int [k];
int j = 0 ;
while (!maxHeap.isEmpty()) {
//Add the remaining elements to the answer
ans[j] = nums[maxHeap.poll().ind];
j++;
}
// reverse the array to get elements closest elements in ascending order
for ( int i = 0 ; i < k / 2 ; i++) {
int t = ans[i];
ans[i] = ans[k - i - 1 ];
ans[k - i - 1 ] = t;
}
return ans;
}
public static void main(String[] args)
{
int arr[] = { 12 , 16 , 22 , 30 , 35 , 39 , 42 ,
45 , 48 , 50 , 53 , 55 , 56 };
int k = 4 , x = 35 ;
int ans[] = printKClosest(arr, arr.length, k, x);
System.out.println(Arrays.toString(ans));
}
} |
# Python Code for the above approach import heapq
def findClosestElements(arr, k, x):
# Create a max heap to store the pairs of absolute differences and negative values
max_heap = []
for num in arr:
# Skip if the element is equal to x
if num = = x:
continue
# Calculate the absolute difference and add the pair to the max heap
diff = abs (num - x)
heapq.heappush(max_heap, ( - diff, num))
# If the size of the max heap exceeds k, remove the element with the maximum absolute difference
if len (max_heap) > k:
heapq.heappop(max_heap)
# Store the result in an array
result = []
# Retrieve the top k elements from the max heap
while max_heap:
# Get the top element from the max heap
diff, num = heapq.heappop(max_heap)
# Add the value to the result array
result.append(num)
# Return the closest numbers in ascending order
return sorted (result)
# Test case arr = [ 12 , 16 , 22 , 30 , 35 , 39 , 42 , 45 , 48 , 50 , 53 , 55 , 56 ]
k = 4
x = 35
res = findClosestElements(arr, k, x)
print (res)
# THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL |
//C# Code for the above approach using System;
using System.Collections.Generic;
using System.Linq;
class Program
{ // Function to find the k closest elements to x in the given array
static List< int > FindClosestElements(List< int > arr, int k, int x)
{
// Sort the array by absolute difference from x and then by value
arr.Sort((a, b) => Math.Abs(a - x).CompareTo(Math.Abs(b - x)) == 0 ? a.CompareTo(b) : Math.Abs(a - x).CompareTo(Math.Abs(b - x)));
// Take the first k elements as they will be the closest
var result = arr.GetRange(0, k);
return result;
}
static void Main()
{
// Test case
List< int > arr = new List< int > { 12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56 };
int k = 4;
int x = 35;
// Remove the value x from the list before finding the closest elements
arr.Remove(x);
List< int > res = FindClosestElements(arr, k, x);
// Print the result
Console.WriteLine( string .Join( " " , res));
}
} |
function findClosestElements(arr, k, x) {
// Create a max heap to store the pairs of absolute
// differences and negative values
const maxH = new MaxHeap();
const n = arr.length;
for (let i = 0; i < n; i++) {
// Skip if the element is equal to x
if (arr[i] === x)
continue ;
// Calculate the absolute difference and add the
// pair to the max heap
const diff = Math.abs(arr[i] - x);
maxH.insert({ diff: diff, value: -arr[i] });
// If the size of the max heap exceeds k, remove the
// element with the maximum absolute difference
if (maxH.size() > k)
maxH.extractMax();
}
// Store the result in an array
const result = [];
// Retrieve the top k elements from the max heap
while (!maxH.isEmpty()) {
// Get the top element from the max heap
const p = maxH.extractMax();
// Add the negative value to the result array
result.push(-p.value);
}
// Reverse the result array to get the closest numbers
// in ascending order
result.reverse();
return result;
} // MaxHeap class implementation class MaxHeap { constructor() {
this .heap = [];
}
size() {
return this .heap.length;
}
isEmpty() {
return this .heap.length === 0;
}
insert(value) {
this .heap.push(value);
this .heapifyUp( this .heap.length - 1);
}
extractMax() {
if ( this .isEmpty())
return null ;
const max = this .heap[0];
const lastElement = this .heap.pop();
if (! this .isEmpty()) {
this .heap[0] = lastElement;
this .heapifyDown(0);
}
return max;
}
heapifyUp(index) {
const parentIndex = Math.floor((index - 1) / 2);
if (parentIndex >= 0 && this .heap[parentIndex].diff <
this .heap[index].diff) {
this .swap(parentIndex, index);
this .heapifyUp(parentIndex);
}
}
heapifyDown(index) {
const leftChildIndex = 2 * index + 1;
const rightChildIndex = 2 * index + 2;
let largestIndex = index;
if (leftChildIndex < this .heap.length &&
this .heap[leftChildIndex].diff >
this .heap[largestIndex].diff) {
largestIndex = leftChildIndex;
}
if (rightChildIndex < this .heap.length &&
this .heap[rightChildIndex].diff >
this .heap[largestIndex].diff) {
largestIndex = rightChildIndex;
}
if (largestIndex !== index) {
this .swap(largestIndex, index);
this .heapifyDown(largestIndex);
}
}
swap(index1, index2) {
[ this .heap[index1], this .heap[index2]] =
[ this .heap[index2], this .heap[index1]];
}
} // Test case const arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]; const k = 4; const x = 35; const res = findClosestElements(arr, k, x); console.log(res); |
39 30 42 45
Time Complexity: O(n log k), where n is the size of the array and k is the number of elements to be returned. The priority queue takes O(log k) time to insert an element and O(log k) time to remove the top element. Therefore, traversing through the array and inserting elements into the priority queue takes O(n log k) time. Popping elements from the priority queue and pushing them into the result vector takes O(k log k) time. Therefore, the total time complexity is O(n log k + k log k) which is equivalent to O(n log k).
Auxiliary Space: O(k), as we are using a priority queue of size k+1 and a vector of size k to store the result.
Exercise: Extend the optimized solution to work for duplicates also, i.e., to work for arrays where elements don’t have to be distinct.
Approach 3: Two Pointer Approach
Initialization:
- Initialize two pointers, left and right, pointing to the start and end of the array, respectively.
Binary Search:
- Use a binary search-like approach to narrow down the range until right – left >= k.
- Compare the absolute differences of the elements at left and right with the target.
- Move the pointers accordingly to minimize the absolute difference.
Print Result:
- Print the k closest elements within the narrowed down range.
#include <iostream> using namespace std;
void findKClosestElements( int nums[], int target, int k, int n)
{ int left = 0;
int right = n;
while (right - left >= k)
{
if ( abs (nums[left] - target) > abs (nums[right] - target)) {
left++;
}
else {
right--;
}
}
// print the `k` closest elements
while (left <= right)
{
cout<<nums[left]<< " " ;;
left++;
}
} int main( void )
{ int nums[] = { 12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56 };
int n = sizeof (nums) / sizeof (nums[0]);
int target = 35, k = 4;
findKClosestElements(nums, target, k, n);
return 0;
} |
import java.util.Arrays;
public class KClosestElements {
// Function to find k closest elements to the target in the array
static void findKClosestElements( int [] nums, int target, int k, int n) {
int left = 0 ;
int right = n - 1 ;
// Iterate until the difference between right and left pointers is >= k
while (right - left >= k) {
// Compare absolute differences between target and elements
// at left and right pointers
// Move the pointers closer to the target by updating left/
// right based on the closest element
if (Math.abs(nums[left] - target) > Math.abs(nums[right] - target)) {
left++;
} else {
right--;
}
}
// Print the k closest elements
while (left <= right) {
System.out.print(nums[left] + " " );
left++;
}
}
public static void main(String[] args) {
int [] nums = { 12 , 16 , 22 , 30 , 35 , 39 , 42 , 45 , 48 , 50 , 53 , 55 , 56 };
int n = nums.length;
int target = 35 , k = 4 ;
// Call the function to find and print k closest elements to the target
findKClosestElements(nums, target, k, n);
}
} |
def find_k_closest_elements(nums, target, k, n):
left = 0
right = n - 1
# Use binary search to find the closest elements
while right - left > = k:
# Compare the absolute differences between the target and elements at left and right pointers
if abs (nums[left] - target) > abs (nums[right] - target):
left + = 1
else :
right - = 1
# Print the `k` closest elements
while left < = right:
print (nums[left], end = " " )
left + = 1
if __name__ = = "__main__" :
nums = [ 12 , 16 , 22 , 30 , 35 , 39 , 42 , 45 , 48 , 50 , 53 , 55 , 56 ]
n = len (nums)
target = 35
k = 4
# Find and print the k closest elements to the target
find_k_closest_elements(nums, target, k, n)
|
using System;
class Program
{ // Function to find and print the K closest elements to the target in the given array
static void FindKClosestElements( int [] nums, int target, int k, int n)
{
// Initialize two pointers, 'left' and 'right'
int left = 0;
int right = n;
// Continue the loop until the difference between 'right' and 'left' is greater than 'k'
while (right - left > k)
{
// Compare the absolute differences of elements at 'left' and 'right - 1' with the target
if (Math.Abs(nums[left] - target) > Math.Abs(nums[right - 1] - target))
{
// If the left element is farther from the target, move 'left' pointer to the right
left++;
}
else
{
// If the right element is farther from the target, move 'right' pointer to the left
right--;
}
}
// Print the `k` closest elements between the 'left' and 'right' pointers
for ( int i = left; i < right; i++)
{
Console.Write(nums[i] + " " );
}
}
// Main method, entry point of the program
static void Main()
{
// Given sorted array
int [] nums = { 12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56 };
int n = nums.Length; // Length of the array
int target = 35, k = 4; // Target value and the number of closest elements to find
// Call the function to find and print the K closest elements
FindKClosestElements(nums, target, k, n);
// ReadLine to keep the console window open until user presses Enter
Console.ReadLine();
}
} // This code is contributed by shivamgupta310570 |
function find_k_closest_elements(nums, target, k, n) {
let left = 0;
let right = n - 1;
// Use binary search to find the closest elements
while (right - left >= k) {
// Compare the absolute differences between the target and elements at left and right pointers
if (Math.abs(nums[left] - target) > Math.abs(nums[right] - target)) {
left++;
} else {
right--;
}
}
// Print the `k` closest elements
while (left <= right) {
console.log(nums[left], end= " " );
left++;
}
} // Driver Code if ( true ) {
const nums = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56];
const n = nums.length;
const target = 35;
const k = 4;
// Find and print the k closest elements to the target
find_k_closest_elements(nums, target, k, n);
} |
30 35 39 42
The time complexity of the above solution is O(n) and doesn’t require any extra space.